Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
T power(T a, T b) {
  T x;
  if (b == 0)
    x = 1;
  else
    x = a;
  for (size_t i = 1; i < b; i++) x *= a;
  return x;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int n, m, a, b, c, ab, ba, x, y, z, avg, sum;
long long int arr2[101][101], arr3[200000], arr4[200001], arr5[100001];
char c2[1001][5];
bool f, f1, f2;
string str[50], s1, s2, s3, s4;
set<int> seto;
double d, d1;
long long int x1, x2, y11, y2;
long long int mini = 100000000000000, maxi = 0;
map<string, long long int> mp;
vector<long long int> v, v1;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cin >> arr2[i][j];
  }
  if (n < m) {
    for (int i = 0; i < n; i++) {
      a = 100000000000000;
      for (int j = 0; j < m; j++) {
        if (arr2[i][j] < a) a = arr2[i][j];
      }
      if (a > 0) {
        for (int j = 0; j < m; j++) {
          arr2[i][j] -= a;
        }
        for (int j = 0; j < a; j++) v.push_back(i + 1);
      }
    }
    for (int i = 0; i < m; i++) {
      a = 100000000000000;
      for (int j = 0; j < n; j++) {
        if (arr2[j][i] < a) a = arr2[j][i];
      }
      if (a > 0) {
        for (int j = 0; j < n; j++) {
          arr2[j][i] -= a;
        }
        for (int j = 0; j < a; j++) v1.push_back(i + 1);
      }
    }
  } else {
    for (int i = 0; i < m; i++) {
      a = 100000000000000;
      for (int j = 0; j < n; j++) {
        if (arr2[j][i] < a) a = arr2[j][i];
      }
      if (a > 0) {
        for (int j = 0; j < n; j++) {
          arr2[j][i] -= a;
        }
        for (int j = 0; j < a; j++) v1.push_back(i + 1);
      }
    }
    for (int i = 0; i < n; i++) {
      a = 100000000000000;
      for (int j = 0; j < m; j++) {
        if (arr2[i][j] < a) a = arr2[i][j];
      }
      if (a > 0) {
        for (int j = 0; j < m; j++) {
          arr2[i][j] -= a;
        }
        for (int j = 0; j < a; j++) v.push_back(i + 1);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (arr2[i][j]) {
        cout << -1;
        return 0;
      }
    }
  }
  cout << v.size() + v1.size() << endl;
  for (int i = 0; i < v.size(); i++) cout << "row " << v[i] << endl;
  for (int i = 0; i < v1.size(); i++) cout << "col " << v1[i] << endl;
}